A099921 a(n) = 5*Fibonacci(n)^2.

5, 5, 20, 45, 125, 320, 845, 2205, 5780, 15125, 39605, 103680, 271445, 710645, 1860500, 4870845, 12752045, 33385280, 87403805, 228826125, 599074580, 1568397605, 4106118245, 10749957120, 28143753125, 73681302245, 192900153620, 505019158605, 1322157322205

Offset: 1

Author

Ralf Stephan, Nov 01 2004

Keywords

• nonn
• easy

References

• Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 53.

Links

• Matthew House, Table of n, a(n) for n = 1..2380
• Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Crossrefs

Cf. A000032, A000045 (Fibonacci numbers), A007598.

Programs

Magma

[5*Fibonacci(n)^2: n in [1..30]]; // Vincenzo Librandi, Sep 14 2017

Maple

A099921:=n->5*combinat[fibonacci](n)^2: seq(A099921(n), n=1..50); # Wesley Ivan Hurt, Sep 16 2017

Mathematica

5*Fibonacci[Range[30]]^2 (* Harvey P. Dale, Feb 24 2015 *)
LinearRecurrence[{2, 2, -1}, {5, 5, 20}, 30] (* Vincenzo Librandi, Sep 14 2017 *)

PARI

a(n) = 5*fibonacci(n)^2; \\ Michel Marcus, Jan 14 2017

PARI

Vec(x*(5-5*x) / ((1+x)*(1-3*x+x^2)) + O(x^40)) \\ Colin Barker, Jan 14 2017

Formula

a(n) = 5 * A007598(n).

a(n) = Lucas(n)^2 - 4(-1)^n.

G.f.: x*(5-5*x) / ((1+x)*(1-3*x+x^2)).

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3. - Matthew House, Jan 13 2017

a(n) = ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n - 2*(-1)^n. - Colin Barker, Jan 14 2017

a(n) = 2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n. - Vincenzo Librandi, Sep 14 2017

a(n) = Lucas(2*n) - 2*(-1)^n. - Amiram Eldar, Jan 17 2022